# Create a Huge Problem

I am wondering if any problems have been designed that test a wide range of mathematical skills. For example, I remember doing the integral $$\int \sqrt{\tan x}\;\mathrm{d}x$$ and being impressed at how many techniques (substitution, trig, partial fractions etc.) I had to use to solve it successfully.

I am looking for suggestions/contributions to help build up such a question. For example, one problem could have as its answer $\tan x$ which would then be used in the integral, and something about the answer to the integral could lead into the next part.

The relevant subjects would be anything covered in the first few years of an undergraduate degree in mathematics.

The main reason I ask is that I want to work on developing some more integrated ways to practice mathematics “holistically” which I feel is very lacking in the current educational model.

I don’t know this is eligible answer or not but I think it’s worth being shared. This solution I made when trying to evaluate
$$\int_0^{\infty}\frac{\cos2x}{x^2+4}\,dx$$
on Brilliant.org. These are the methods I use to evaluate that integral and post it there as a solution.

Method 1:

Consider the function $$f(t)=e^{-a|t|}$$, then the Fourier transform of $$f(t)$$ is given by
\begin{align} F(\omega)=\mathcal{F}[f(t)]&=\int_{-\infty}^{\infty}f(t)e^{-i\omega t}\,dt\\ &=\int_{-\infty}^{\infty}e^{-a|t|}e^{-i\omega t}\,dt\\ &=\int_{-\infty}^{0}e^{at}e^{-i\omega t}\,dt+\int_{0}^{\infty}e^{-at}e^{-i\omega t}\,dt\\ &=\lim_{u\to-\infty}\left. \frac{e^{(a-i\omega)t}}{a-i\omega} \right|_{t=u}^0-\lim_{v\to\infty}\left. \frac{e^{-(a+i\omega)t}}{a+i\omega} \right|_{0}^{t=v}\\ &=\frac{1}{a-i\omega}+\frac{1}{a+i\omega}\\ &=\frac{2a}{\omega^2+a^2}. \end{align}
Next, the inverse Fourier transform of $$F(\omega)$$ is
\begin{align} f(t)=\mathcal{F}^{-1}[F(\omega)]&=\frac{1}{2\pi}\int_{-\infty}^{\infty}F(\omega)e^{i\omega t}\,d\omega\\ e^{-a|t|}&=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{2a}{\omega^2+a^2}e^{i\omega t}\,d\omega\\ \frac{\pi e^{-a|t|}}{a}&=\int_{-\infty}^{\infty}\frac{e^{i\omega t}}{\omega^2+a^2}\,d\omega.\tag1 \end{align}
Now, rewrite
$$\int_0^{\infty}\frac{\cos2x}{x^2+4}\,dx=\frac{1}{2}\int_{-\infty}^{\infty}\frac{\mathbb{Re}\left(e^{2ix}\right)}{x^2+2^2}\,dx.\tag2$$
Comparing $$(2)$$ to $$(1)$$ yield $$t=2$$ and $$a=2$$. Thus,
\begin{align} \int_0^{\infty}\frac{\cos2x}{x^2+4}\,dx &=\frac{1}{2}\frac{\pi e^{-2\cdot|2|}}{2}\\ &=\frac{\pi}{4e^4}\\ \end{align}

Method 2:

Note that:
$$\int_{y=0}^\infty e^{-(x^2+4)y}\,dy=\frac{1}{x^2+4},$$
therefore
$$\int_{x=0}^\infty\int_{y=0}^\infty e^{-(x^2+4)y}\cos2x\,dy\,dx=\int_0^{\infty}\frac{\cos2x}{x^2+4}\,dx$$
Rewrite $$\cos2x=\Re\left(e^{-2ix}\right)$$, then
\begin{align} \int_0^{\infty}\frac{\cos2x}{x^2+4}\,dx&=\frac{1}{2}\int_{-\infty}^{\infty}\frac{\cos2x}{x^2+4}\,dx\\ &=\frac{1}{2}\int_{-\infty}^{\infty}\int_{y=0}^\infty e^{-(x^2+4)y}\cos2x\,dy\,dx\\ &=\frac{1}{2}\int_{y=0}^\infty\int_{x=-\infty}^\infty e^{-(yx^2+2ix+4y)}\,dx\,dy\\ &= \int_{y=0}^\infty e^{-4y} \int_{x=-\infty}^\infty \frac{1}{2} e^{-(yx^2+2ix)}\,dx\,dy. \end{align}
In general
\begin{align} \int_{x=-\infty}^\infty e^{-(ax^2+bx)}\,dx&=\int_{x=-\infty}^\infty \exp\left(-a\left(\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a^2}\right)\right)\,dx\\ &=\exp\left(\frac{b^2}{4a}\right)\int_{x=-\infty}^\infty \exp\left(-a\left(x+\frac{b}{2a}\right)^2\right)\,dx\\ \end{align}
Let $$u=x+\frac{b}{2a}\;\rightarrow\;du=dx$$, then
\begin{align} \int_{x=-\infty}^\infty e^{-(ax^2+bx)}\,dx&=\exp\left(\frac{b^2}{4a}\right)\int_{x=-\infty}^\infty \exp\left(-a\left(x+\frac{b}{2a}\right)^2\right)\,dx\\ &=\exp\left(\frac{b^2}{4a}\right)\int_{u=0}^\infty e^{-au^2}\,du.\\ \end{align}
The last form integral is Gaussian integral that equals to $$\sqrt{\dfrac{\pi}{a}}$$. Hence
$$\int_{x=-\infty}^\infty e^{-(ax^2+bx)}\,dx=\sqrt{\frac{\pi}{a}}\exp\left(\frac{b^2}{4a}\right).$$
Thus
$$\frac{1}{2}\int_{x=-\infty}^\infty e^{-(yx^2+2ix)}\,dx=\frac{1}{2}\sqrt{\frac{\pi}{y}}\exp\left(\frac{(2i)^2}{4y}\right)=\frac{1}{2}\sqrt{\frac{\pi}{y}}\exp\left(-\frac{1}{y}\right).$$
Next
$$\int_0^{\infty}\frac{\cos2x}{x^2+4}\,dx=\frac{\sqrt{\pi}}{2}\int_{y=0}^\infty \frac{\exp\left(-4y-\frac{1}{y}\right)}{\sqrt{y}}\,dy.$$
In general
\begin{align} \int_{y=0}^\infty \frac{\exp\left(-ay-\frac{b}{y}\right)}{\sqrt{y}}\,dy&=2\int_{v=0}^\infty \exp\left(-av^2-\frac{b}{v^2}\right)\,dv\\ &=2\int_{v=0}^\infty \exp\left(-a\left(v^2+\frac{b}{av^2}\right)\right)\,dv\\ &=2\int_{v=0}^\infty \exp\left(-a\left(v^2-2\sqrt{\frac{b}{a}}+\frac{b}{av^2}+2\sqrt{\frac{b}{a}}\right)\right)\,dv\\ &=2\int_{v=0}^\infty \exp\left(-a\left(v-\frac{1}{v}\sqrt{\frac{b}{a}}\right)^2-2\sqrt{ab}\right)\,dv\\ &=2\exp(-2\sqrt{ab})\int_{v=0}^\infty \exp\left(-a\left(v-\frac{1}{v}\sqrt{\frac{b}{a}}\right)^2\right)\,dv\\ \end{align}
The trick to solve the last integral is by setting
$$I=\int_{v=0}^\infty \exp\left(-a\left(v-\frac{1}{v}\sqrt{\frac{b}{a}}\right)^2\right)\,dv.$$
Let $$t=-\frac{1}{v}\sqrt{\frac{b}{a}}\;\rightarrow\;v=-\frac{1}{t}\sqrt{\frac{b}{a}}\;\rightarrow\;dv=\frac{1}{t^2}\sqrt{\frac{b}{a}}\,dt$$, then
$$I_t=\sqrt{\frac{b}{a}}\int_{t=0}^\infty \frac{\exp\left(-a\left(-\frac{1}{t}\sqrt{\frac{b}{a}}+t\right)^2\right)}{t^2}\,dt.$$
Let $$t=v\;\rightarrow\;dt=dv$$, then
$$I_t=\int_{t=0}^\infty \exp\left(-a\left(t-\frac{1}{t}\sqrt{\frac{b}{a}}\right)^2\right)\,dt.$$
Adding the two $$I_t$$s yields
$$2I=I_t+I_t=\int_{t=0}^\infty\left(1+\frac{1}{t^2}\sqrt{\frac{b}{a}}\right)\exp\left(-a\left(t-\frac{1}{t}\sqrt{\frac{b}{a}}\right)^2\right)\,dt.$$
Let $$s=t-\frac{1}{t}\sqrt{\frac{b}{a}}\;\rightarrow\;ds=\left(1+\frac{1}{t^2}\sqrt{\frac{b}{a}}\right)dt$$ and for $$0 is corresponding to $$-\infty, then
$$I=\frac{1}{2}\int_{s=-\infty}^\infty e^{-as^2}\,ds=\frac{1}{2}\sqrt{\frac{\pi}{a}}.$$
Thus
\begin{align} \int_{y=0}^\infty \frac{\exp\left(-ay-\frac{b}{y}\right)}{\sqrt{y}}\,dy&=2\exp(-2\sqrt{ab})\int_{v=0}^\infty \exp\left(-a\left(v-\frac{1}{v}\sqrt{\frac{b}{a}}\right)^2\right)\,dv\\ &=\sqrt{\frac{\pi}{a}}e^{-2\sqrt{ab}}\\ \end{align}
and
\begin{align} \int_0^{\infty}\frac{\cos2x}{x^2+4}\,dx&=\frac{\sqrt{\pi}}{2}\int_{y=0}^\infty \frac{\exp\left(-4y-\frac{1}{y}\right)}{\sqrt{y}}\,dy\\ &=\frac{\sqrt{\pi}}{2}\cdot\sqrt{\frac{\pi}{4}}e^{-2\sqrt{4\cdot1}}\\ &=\frac{\pi}{4e^4}. \end{align}
It took me hours to make a solution using method 2 because almost no one on that site knows Fourier transform (it is understandable since most of the user there are only school students, so they know nothing about this method) so I made a solution using standard methods. Other methods to solve it is using contour integral or residual theorem, but I’m not familiar with those methods.